Isolating x in Equation: 2x\sqrt{3+\sqrt{7}}=\sqrt{(x+1)^2+((\sqrt{7}+2)x-1)^2}

[Mentallic]

How could I possibly go about isolating x in the following equation:

2x\sqrt{3+\sqrt{7}}=\sqrt{(x+1)^2+((\sqrt{7}+2)x-1)^2}-2+2\sqrt{2}

Any suggestions? The problem I have here is that squaring the entire equation only leads to more trouble. Any help is appreciated

 

[Tac-Tics]

How could I possibly go about isolating x in the following equation:

2x\sqrt{3+\sqrt{7}}=\sqrt{(x+1)^2+((\sqrt{7}+2)x-1)^2}-2+2\sqrt{2}

Any suggestions? The problem I have here is that squaring the entire equation only leads to more trouble. Any help is appreciated

Why are you dealing with such a monstrocity?

I didn't go through all the algebra, but I'm not sure you can isolate x on one side. The worst case is that you end up with a polynomial of degree 4.

If you're serious about trying to solve for x, my hint would be group all the constant expressions together and give them variable names so you can focus on getting the x's isolated without having to worry about whether or not you correctly squared (sqrt(7)+2).

So let A = 2 sqrt(3 + sqrt(7)), B = sqrt(7) + 2, C = -2 + 2 sqrt(2).
Then write the problem as Ax = sqrt( (x+1)^2 + (Bx - 1)^2 ) + C.
Ax - C = sqrt( (x+1)^2 + (Bx - 1)^2 )
(Ax - C)^2 = (x+1)^2 + (Bx - 1)^2

You can do the rest.

 

[Mentallic]

Why are you dealing with such a monstrocity?

This is the last part (I think) to answering https://www.physicsforums.com/showthread.php?t=265300" question and it only took me 4 months to do so (I took a nice break).

I didn't go through all the algebra, but I'm not sure you can isolate x on one side. The worst case is that you end up with a polynomial of degree 4.

Glad I wasn't left dealing with the worst case scenario then!

If you're serious about trying to solve for x, my hint would be group all the constant expressions together and give them variable names so you can focus on getting the x's isolated without having to worry about whether or not you correctly squared (sqrt(7)+2).

So let A = 2 sqrt(3 + sqrt(7)), B = sqrt(7) + 2, C = -2 + 2 sqrt(2).
Then write the problem as Ax = sqrt( (x+1)^2 + (Bx - 1)^2 ) + C.
Ax - C = sqrt( (x+1)^2 + (Bx - 1)^2 )
(Ax - C)^2 = (x+1)^2 + (Bx - 1)^2

You can do the rest.

That was a nice suggestion. And while it made things much simpler to work with the variables rather than combos of surds, I end up with a fairly complicated equation nonetheless (and this scares me even more knowing I have to sub back!).

I am left with a quadratic in x:

x^2(A^2-B^2-1)+2x(B-AC-1)-C^2-2=0

Now my only problem is simplifying - as much as possible - the equation of pronumerals that are meshed up in the quadratic formula.
The furthest I can seem to get to is:

x=\frac{AC-B+1 \pm \sqrt{(1+C^2)(2A^2-B^2-1)-2B(1+AC)+2AC}}{A^2-B^2-1}

Now for the substituting and simplifying from there.

 

[gabbagabbahey]

How could I possibly go about isolating x in the following equation:

2x\sqrt{3+\sqrt{7}}=\sqrt{(x+1)^2+((\sqrt{7}+2)x-1)^2}-2+2\sqrt{2}

Any suggestions? The problem I have here is that squaring the entire equation only leads to more trouble. Any help is appreciated

Assuming that x represents the x-coordinate of the center of the circle you are trying to find, then this equation cannot possibly be correct.

 

[Mentallic]

Assuming that x represents the x-coordinate of the center of the circle you are trying to find, then this equation cannot possibly be correct.

Yes I made a slight error in that part. But there were more major errors that wouldn't show themselves to me until after an hour of searching.

The equation should have been:

2x\sqrt{3+\sqrt{7}}-1=\sqrt{(x+1)^2+((\sqrt{7}+2)x-1)^2}-\sqrt{3-2\sqrt{2}}

Eventually, after much squandering, I solved the question:

The circle required is:

(x+A)^2+(y+(\sqrt{7}+2)A)^2=(2\sqrt{3+\sqrt{7}}A-1)^2

where A=\frac{2(1-\sqrt{2})}{1+\sqrt{7}+2\sqrt{3+\sqrt{7}}(\sqrt{2}-2)}

or, more approximately:

(x-0.96)^2+(y-4.465)^2=12.724

 

[gabbagabbahey]

Yes I made a slight error in that part. But there were more major errors that wouldn't show themselves to me until after an hour of searching.

The equation should have been:

2x\sqrt{3+\sqrt{7}}-1=\sqrt{(x+1)^2+((\sqrt{7}+2)x-1)^2}-\sqrt{3-2\sqrt{2}}

Eventually, after much squandering, I solved the question:

The circle required is:

(x+A)^2+(y+(\sqrt{7}+2)A)^2=(2\sqrt{3+\sqrt{7}}A-1)^2

where A=\frac{2(1-\sqrt{2})}{1+\sqrt{7}+2\sqrt{3+\sqrt{7}}(\sqrt{2}-2)}

or, more approximately:

(x-0.96)^2+(y-4.465)^2=12.724

That's one solution; but there's also another

 

[Mentallic]

That's one solution; but there's also another

But there can't be. I'll post a graphical representation and you will see why (and why I won't bother searching through my pages of notes to find where I went wrong).
I'm curious though, what made you think/assume there was a 2nd solution?

http://img502.imageshack.us/img502/913/circleshy5.jpg
http://g.imageshack.us/img502/circleshy5.jpg/1/

 

[gabbagabbahey]

I'm curious though, what made you think/assume there was a 2nd solution?

This did:

http://img240.imageshack.us/img240/2162/circlesiy8.jpg

 

[Mentallic]

Ahh I see where the 2nd circle is going to be. It will have its centre somewhere in the unit circle and touching the other end of the circles once. My calculations have been restricted to the one circle. This one is going to be considerably harder to solve

Gabbagabbahey, I'm still trying to churn through the help you provided in the original thread. I'll see what I can conclude from it.

Edit: While the picture you posted isn't working, I have a feeling I know what it's going to look like